Recent content by IvanSaurus

Now I am totally lost :) a guess would be "0".. If the Gaussian surface is the purple ring, and we add up Q/2 - Q/2. It still confuses me that outer surface of the conductor reaches 3/2 Q

Well that exactly what I can't seem to figure out. I must have overseen some rule regarding Gaussian Surfaces- since appearantly the charge triples outside the surface? from Q/2 to 2Q/3. The Q/2 is the enclosed charge..

Ill try to wrap my head around it some more later, thanks for the answer though :)

The problem just states that the center of gaussian sphere has charge enclosed, Q/2.. no real values are given.

Hi there, I have following problem at hand: (any help would be appreaciated, I have a test tomorrow) 1. Homework Statement Homework Equations The Attempt at a Solution Above is obviously the teachers answer to the problem, what I don't get though; is, where does the 3/2 come from, at r...